1. ## trig function derivative

Given that $\displaystyle x = cosy^{2}$

find dy/dx in terms of y;

I am not sure which rule to use, do I use the product rule or chain rule?

I thought I should use the chain rule and this is what I get;

first differentiate cos which = -sin

than do y^2 to get 2y.

so $\displaystyle \frac{dx}{dy} = -sin2y$

I know this wrong, but can someone please tell me what I am doing wrong?
the correct answer is $\displaystyle \frac{dx}{dy} = -siny^{2} \times 2y$

thanks,

2. Originally Posted by Tweety
Given that $\displaystyle x = cosy^{2}$

find dy/dx in terms of y;

I am not sure which rule to use, do I use the product rule or chain rule?

I thought I should use the chain rule and this is what I get;

first differentiate cos which = -sin

than do y^2 to get 2y.

so $\displaystyle \frac{dx}{dy} = -sin2y$

I know this wrong, but can someone please tell me what I am doing wrong?
the correct answer is $\displaystyle \frac{dx}{dy} = -siny^{2} \times 2y$

thanks,
which do you mean ...

$\displaystyle x = \cos(y^2)$ or $\displaystyle x = (\cos{y})^2$

?

3. Originally Posted by skeeter
which do you mean ...

$\displaystyle x = \cos(y^2)$ or $\displaystyle x = (\cos{y})^2$

?
The first one. $\displaystyle x = \cos(y^2)$

4. Originally Posted by Tweety
The first one. $\displaystyle x = \cos(y^2)$
your question says to find $\displaystyle \frac{dy}{dx}$, but the answer is $\displaystyle \frac{dx}{dy}$ , so I'll do both.

$\displaystyle \frac{d}{dx}[x = \cos(y^2)]$

chain rule on the right side ...

$\displaystyle 1 = -\sin(y^2) \cdot 2y \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = -\frac{1}{2y\sin(y^2)}$

$\displaystyle \frac{d}{dy}[x = \cos(y^2)]$

$\displaystyle \frac{dx}{dy} = -\sin(y^2) \cdot 2y$

$\displaystyle \frac{dx}{dy} = -2y \sin(y^2)$

5. I think am not understanding how your using the chain rule on $\displaystyle cosy^{2}$

I use this expression ;
$\displaystyle y = f(x)^{n} = \frac{dy}{dx} = nf(x)^{n-1} f '(x)$

there's no 'inside bit in the expression $\displaystyle cosy^{2}$ its just one, so how do you separate the variables?

for example in this expression $\displaystyle y = 6(4x-2)^{3}$ I would first differenciate the expression in the bracket and than do the out side leaving the inside the same and than multiply both expressions.

So am just wondering how $\displaystyle cosy^{2}$ fits into the above formula.

Thank you.

6. note that for $\displaystyle x = \cos(y^2)$ , only the y is squared, not the cosine function.

the derivative w/r to x is

$\displaystyle 1 = -\sin(y^2) \cdot (the \, derivative \, of \, y^2) = -sin(y^2) \cdot 2y \cdot \frac{dy}{dx}$